= Solution
{c}
Define the classical axial spin by the <axial angular velocity of a symmetric top>,
$$
\boxed{n=\boldsymbol\omega\cdot\mathbf e.}
$$
Since $\mathbf e$ is a <unit vector>, $\mathbf e\cdot\dot{\mathbf e}=0$. The <vector triple product> identity applied to $\dot{\mathbf e}=\boldsymbol\omega\wedge\mathbf e$ gives
$$
\mathbf e\wedge\dot{\mathbf e}
=\boldsymbol\omega-(\boldsymbol\omega\cdot\mathbf e)\mathbf e,
\qquad
\boldsymbol\omega=\mathbf e\wedge\dot{\mathbf e}+n\mathbf e.
$$
The <inertia tensor> multiplies the component perpendicular to $\mathbf e$ by $A$ and the component along $\mathbf e$ by $C$. Therefore the <angular momentum> about $Q$ is
$$
\mathbf L_Q=A\mathbf e\wedge\dot{\mathbf e}+Cn\mathbf e.
$$
Its time <derivative> is
$$
\dot{\mathbf L}_Q=A\mathbf e\wedge\ddot{\mathbf e}+C\dot n\mathbf e+Cn\dot{\mathbf e},
$$
because $\dot{\mathbf e}\wedge\dot{\mathbf e}=\mathbf0$.
The gravitational <torque> about $Q$ is $h\mathbf e\wedge(-mg\mathbf k)$, and the applied <torque> is $\ell\mathbf e\wedge(\kappa\mathbf s)$. The support <force> acts at $Q$, so has zero <torque> about $Q$. The <angular momentum> equation is consequently
$$
\boxed{A\mathbf e\wedge\ddot{\mathbf e}+C\dot n\mathbf e+Cn\dot{\mathbf e}
=-mgh\mathbf e\wedge\mathbf k+\kappa\ell\mathbf e\wedge\mathbf s.}
$$
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